Divide the following complex numbers. $ \dfrac{8-2i}{3-5i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${3+5i}$ $ \dfrac{8-2i}{3-5i} = \dfrac{8-2i}{3-5i} \cdot \dfrac{{3+5i}}{{3+5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(8-2i) \cdot (3+5i)} {(3-5i) \cdot (3+5i)} = \dfrac{(8-2i) \cdot (3+5i)} {3^2 - (-5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(8-2i) \cdot (3+5i)} {(3)^2 - (-5i)^2} = $ $ \dfrac{(8-2i) \cdot (3+5i)} {9 + 25} = $ $ \dfrac{(8-2i) \cdot (3+5i)} {34} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({8-2i}) \cdot ({3+5i})} {34} = $ $ \dfrac{{8} \cdot {3} + {-2} \cdot {3 i} + {8} \cdot {5 i} + {-2} \cdot {5 i^2}} {34} $ Evaluate each product of two numbers. $ \dfrac{24 - 6i + 40i - 10 i^2} {34} $ Finally, simplify the fraction. $ \dfrac{24 - 6i + 40i + 10} {34} = \dfrac{34 + 34i} {34} = 1+i $